Find the area of a triangle ABC, where AB = 6.8m, BC = 9.5m and AC = 10.2m.

Expert Answers
jeew-m eNotes educator| Certified Educator

The cosine of a triangle is given by;

`cosA = (b^2+c^2-a^2)/(2bc)`

A is the angle between AB and AC.

a = Opposite leg to the angle A.(leg BC) = 9.5cm

b = Opposite leg to the angle B.(leg AC) = 10.2cm

c = Opposite leg to the angle C.(leg AB) = 6.8cm


`cosA = (10.2^2+6.8^2-9.5^2)/(2*10.2*6.8) = 0.433`


We know that ` sin^2A+cos^2A = 1`


`sinA = sqrt(1-cos^2A) = sqrt(1-0.433^2) = 0.813`


If we assume that the perpendicular drawn from point B will meet AC at point D;

`AD = ABsinA = 6.8*0.813 = 5.528`


Area of triangle

= 1/2(base length*perpendicular height)

`= 1/2*AC*AD`

`= 1/2*10.2*5.528`

`= 28.193cm^2`


So the area of the triangle is `28.193cm^2`




lemjay eNotes educator| Certified Educator

To find the area of the triangle, given the length of the three sides, we may use the Heron's Formula.

`A = sqrt(s(s-a)(s-b)(s-c))`

where       A -  Area of triangle

         a, b, c - length of  the sides of the triangle

                s - semi-perimeter, `s = (a+b+c)/2`

So, let a =AB=6.8m, b = BC=9.5m and c=AC=10.2m. 

Then, solve for s.

`s = (a+b+c)/2 = (6.8 + 9.5+10.2)/2 = 26.5/2= 13.25`

And, substitute values of , b, c and s to the formula of area of triangle.





Hence, the area of triangle is `31.26m^2` .